Tenth Grade Mathematics: Criteria for Similarity: AAA, SAS, and SSS

Proving that two general polygons are similar requires establishing that all pairs of corresponding angles are equal and all pairs of corresponding sides are proportional. For triangles, however, Euclidean geometry provides powerful similarity criteria: proving that a minimal subset of geometric conditions holds is sufficient to guarantee that the remaining conditions follow automatically. In this chapter, we study and prove the three master criteria for triangle similarity: the Angle-Angle-Angle (AAA) criterion, the Side-Side-Side (SSS) criterion, and the Side-Angle-Side (SAS) criterion.

The AAA (Angle-Angle-Angle) Similarity Criterion

Theorem (AAA Similarity Criterion):
If in two triangles, corresponding angles are equal, then their corresponding
sides are in the same ratio (proportional), and hence the two triangles are similar.

Geometric Verification Scheme:
               A                                  D
              / \                                /              /   \                              /               /_____\ P                          /_____           B       C                          /                                                    /_________                                            E           F

Given: In triangle ABC and triangle DEF, Angle A = Angle D, Angle B = Angle E, and Angle C = Angle F. Construction: On the larger triangle DEF, cut off segment DP equal to AB (DP = AB) and segment DQ equal to AC (DQ = AC). Join PQ.

Proof: In triangle ABC and triangle DPQ: AB = DP (by construction), Angle A = Angle D (given), and AC = DQ (by construction). Therefore, triangle ABC is congruent to triangle DPQ (by SAS Congruence). This congruence implies that Angle B = Angle DPQ (corresponding parts of congruent triangles). However, it was given that Angle B = Angle E. Therefore, Angle DPQ = Angle E. Because these corresponding angles are equal, the line segment PQ must be parallel to EF (PQ || EF). By the Basic Proportionality Theorem in triangle DEF: DP / DE = DQ / DF. Substituting DP = AB and DQ = AC gives: AB / DE = AC / DF. By repeating this construction along side EF, we obtain: AB / DE = BC / EF. Therefore: AB / DE = BC / EF = AC / DF. Since corresponding angles are equal and corresponding sides are proportional, triangle ABC is similar to triangle DEF (triangle ABC ~ triangle DEF).

The AA (Angle-Angle) Corollary

Because the sum of the three interior angles in any triangle is always strictly 180 degrees, if two angles of one triangle are equal to two angles of another triangle, their third angles must be equal:

Corollary (AA Criterion):
If two angles of one triangle are respectively equal to two angles of another
triangle, then the two triangles are similar.

The AA criterion is the most frequently used similarity tool in practical geometric problem solving.

The SSS (Side-Side-Side) Similarity Criterion

Theorem (SSS Similarity Criterion):
If in two triangles, the sides of one triangle are proportional to the sides of
the other triangle, then their corresponding angles are equal, and hence the
two triangles are similar.

Mathematical Formulation:
If  AB / DE = BC / EF = CA / FD,
Then Angle A = Angle D,  Angle B = Angle E,  and  Angle C = Angle F.
Therefore, triangle ABC ~ triangle DEF.

The SAS (Side-Angle-Side) Similarity Criterion

Theorem (SAS Similarity Criterion):
If one angle of a triangle is equal to one angle of another triangle and the sides
including these angles are proportional, then the two triangles are similar.

Condition Check:
In triangle ABC and triangle DEF:
1. Angle A = Angle D  (Included angle)
2. AB / DE = AC / DF  (Including sides are proportional)
Conclusion: triangle ABC ~ triangle DEF.

CRITICAL CAUTION: The equal angle must be the included angle (the angle formed between the two proportional sides). If a non-included angle is equal, similarity cannot be inferred.

Summary Comparison of Similarity Criteria

+-----------+-----------------------------------+-----------------------------------+
| Criterion | Required Given Conditions         | Inferred Deductions               |
+-----------+-----------------------------------+-----------------------------------+
| AA / AAA  | 2 (or 3) corresponding angles are | Sides are proportional:           |
|           | equal: Angle A = D, Angle B = E   | AB/DE = BC/EF = AC/DF             |
+-----------+-----------------------------------+-----------------------------------+
| SSS       | All 3 side ratios are equal:      | Angles are equal:                 |
|           | AB/DE = BC/EF = AC/DF             | Angle A = D, Angle B = E, C = F   |
+-----------+-----------------------------------+-----------------------------------+
| SAS       | 1 angle equal AND including sides | Remaining sides proportional      |
|           | proportional: Angle A=D, AB/DE=AC/DF| and remaining angles equal      |
+-----------+-----------------------------------+-----------------------------------+

Chapter Exercises

Exercise 1: In a figure, line segment AB is parallel to CD, and diagonals AD and BC intersect at point O. Using a similarity criterion for two triangles, prove that OA / OD = OB / OC.

Exercise 2: A vertical pole of length 6 m casts a shadow 4 m long on the ground, and at the same time a tower casts a shadow 28 m long. Find the height of the tower using triangle similarity.

Exercise 3: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a triangle similarity criterion, show that OA / OC = OB / OD.

Solutions and Explanations

Solution 1: Consider triangle AOB and triangle DOC. We are given that AB || CD. Transversal AD creates alternate interior angles: Angle OAB = Angle ODC. Transversal BC creates alternate interior angles: Angle OBA = Angle OCD. Furthermore, the vertically opposite angles at O are equal: Angle AOB = Angle DOC. Therefore, by the AAA (or AA) similarity criterion, triangle AOB is similar to triangle DOC (triangle AOB ~ triangle DOC). Because corresponding sides of similar triangles are in the same ratio: OA / OD = OB / OC = AB / DC. Hence, OA / OD = OB / OC.

Solution 2: Let AB be the vertical pole of height 6 m, and BC be its shadow of length 4 m. Let PQ be the vertical tower of unknown height h m, and QR be its shadow of length 28 m. Since both the pole and the tower are vertical to the horizontal ground, Angle B = Angle Q = 90 degrees. Furthermore, because both shadows are cast at the exact same instant of time, the angle of elevation of the sun is identical in both cases: Angle C = Angle R. By the AA similarity criterion, right triangle ABC is similar to right triangle PQR (triangle ABC ~ triangle PQR). Since corresponding sides are proportional: AB / PQ = BC / QR ==> 6 / h = 4 / 28. Simplify the ratio: 4 / 28 = 1 / 7. Thus: 6 / h = 1 / 7 ==> h = 6 * 7 = 42 m. The height of the tower is 42 meters.

Solution 3: In trapezium ABCD, AB || DC, and the diagonals AC and BD intersect at O. Consider triangle OAB and triangle OCD. Since AB || DC, alternate interior angles formed by transversal AC are equal: Angle OAB = Angle OCD. Alternate interior angles formed by transversal BD are equal: Angle OBA = Angle ODC. Vertically opposite angles are equal: Angle AOB = Angle COD. By the AAA similarity criterion, triangle OAB is similar to triangle OCD (triangle OAB ~ triangle OCD). Because corresponding sides of similar triangles are proportional, the ratio of their corresponding sides is equal: OA / OC = OB / OD.